incomingTheory.tex

Product:
an assessment instrument/list of questions for incoming to algorithms,
in which we expect to see proofs about resource utilization

Recall proof by mathematic induction:\\
There is a proposition, $p$, that we want to show to be true for some domain (i.e., set) $D$.
There is some correspondence between the elements of $D$ and the natural numbers.
When the corresponding natural number, k, is at some value, say $k_0$ for which we can readily show that p is true, we take advantage of that ease, and prove $p(k_0)$, that is to say, prove the proposition p is true for the case $k_0$.\\
Next we need to have a link, from one case to another. Consider a sequence, i.e., the idea "next" makes sense. Then, the implication "$p$ is true of k implies that $p$ is true of next(k) provides a link. We have a successor function for the natural numbers; the next natural number is obtained by adding 1. Consider the implication "$p$ is true of k implies that $p$ is true of $k+1$. 
We can combine a base case, $p(k_0)$ with a linking implication, $p(k) \implies p(k+1)$, where the context in which the linking implication is true includes $k_0$, to show that $p$ is true of all the elements of the domain that are greater than or equal to $k_0$.
The above considerations make useful the following two part procedure.
Find a base case and prove it.
Find a linking implication and prove it, for a domain including the base case.


\begin{enumerate} 

\item  Proofs of Termination
\begin{enumerate}
\item for ($i=0;i<10;i++$)\{\\
    disp(i);\\
\}\\
Assuming there is no problem with disp, how do we prove that this loop terminates?
\item for ($i=0;i>10;i--$)\{\\
    disp(i);\\
\}\\
Assuming there is no problem with disp, how do we prove that this loop terminates?
\item for ($i=n;i>0;i--$)\{\\
    disp(i);\\
\}\\
Assuming there is no problem with disp, how do we prove that this loop terminates?
\item 
i=n;
while ($i>n$)\{\\
    disp(i);\\
    i--;
\}\\
Assuming there is no problem with disp, how do we prove that this loop terminates?
\item 
i=n;
while ($<some\;\;condition>$)\{\\
    disp(i);\\
    update(condition);
\}\\
Assuming there is no problem with disp or update, how do we prove that this loop terminates?
\end{enumerate}

\item Proofs by Loop Invariant \\
Consider the following situation:\\
There is a proposition, $p$, that we want to show to be true for some domain (i.e., set) $D$, \textbf{and we will process elements of that domain using a loop}.
There is some correspondence between the elements of $D$ and the natural numbers.
When the corresponding natural number, k, is at some value, say $k_0$ for which we can readily show that p is true, we take advantage of that ease, and prove $p(k_0)$, that is to say, prove the proposition p is true for the case $k_0$. \textbf{We will choose our loop variable's first value to be $k_0$.}\\
Next we need to have a link, from one case to another. Consider a sequence, i.e., the idea "next" makes sense. Then, the implication "$p$ is true of k implies that $p$ is true of next(k) provides a link. We have a successor function for the natural numbers; the next natural number is obtained by adding 1. Consider the implication "$p$ is true of k implies that $p$ is true of \textbf{next($k$). The loop specification produces a sequence, defining "next".}
We can combine a base case, $p(k_0)$ with a linking implication, $p(k) \implies p($next($k)$, where the context in which the linking implication is true includes $k_0$, to show that $p$ is true of all the elements of the domain that are \textbf{next after} or equal to $k_0$.
The above considerations make useful the following two part procedure.
Find a base case and prove it.
Find a linking implication and prove it, for a domain including the base case.

In the specific situation of proof by loop invariant, these procedure steps are called:\\
Prove that the property is true before the loop starts.
Prove that what happens in the loop does not, at the end of any single loop iteration, change the property to false.

\item 

\end{enumerate}